A200516 Least m>0 such that n = y^2 - 6^x (mod m) has no solution, or 0 if no such m exists.

0, 5, 5, 0, 7, 9, 5, 5, 0, 0, 0, 5, 5, 0, 9, 0, 5, 5, 7, 0, 9, 5, 5, 9, 0, 7, 5, 5, 0, 9, 0, 5, 5, 36, 16, 0, 5, 5, 9, 7, 0, 5, 5, 0, 21, 0, 5, 5, 0, 43, 9, 5, 5, 7, 16, 16, 5, 5, 0, 9, 7, 5, 5, 0, 0, 9, 5, 5, 9, 36, 16, 5, 5, 0, 7, 0, 5, 5, 32, 24, 0, 5, 5

Offset: 0

Author

M. F. Hasler, Nov 18 2011

Keywords

• nonn

Comments

To prove that an integer n is in A051208, it is sufficient to find integers x,y such that y^2 - 6^x = n. In that case, a(n)=0. To prove that n is *not* in A051208, it is sufficient to find a modulus m for which the (finite) set of all possible values of 6^x and y^2 (mod m) allows us to deduce that y^2 - 6^x can never equal n. The present sequence lists the smallest such m>0, if it exists.

Examples

			See A200512 for motivation and detailed examples.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A051204-A051221, A200505-A200520.

Programs

PARI

A200516(n,b=6,p=3)={ my( x=0, qr, bx, seen ); for( m=3,9e9, while( x^p < m, issquare(b^x+n) & return(0); x++); qr=vecsort(vector(m,y,y^2-n)%m,,8); seen=0; bx=1; until( bittest(seen+=1<bx & break; next(3))); return(m))}